Respuesta :
Explanation:
(a) The given data is as follows.
Length of the rod, L = 0.83 m
Mass of the rod, m = 110 g = 0.11 (as 1 kg = 1000 g)
At the lowest point, angular speed of the rod ([tex]\omega[/tex]) = 5.71 rad/s
First, we will calculate the rotational inertia of the rod about an axis passing through its fixed end is as follows.
I = [tex]I_{CM} + mh^{2}[/tex]
= [tex]\frac{1}{12}mL^{2} + m(\frac{L}{2})[/tex]
= [tex]\frac{1}{12} \times 0.11 \times (0.83)^{2} + 0.11 \times \frac{0.83}{2}[/tex]
= 0.00631 + 0.415
= 0.42131 [tex]kg m^{2}[/tex]
Therefore, kinetic energy of the rod at its lowest point will be calculated as follows.
K = [tex]\frac{1}{2}I \omega^{2}[/tex]
= [tex]\frac{1}{2} \times 0.42131 kg m^{2} \times (5.71)^{2}[/tex]
= 6.86 J
Hence, kinetic energy of the rod at its lowest point is 6.86 J.
(b) According to the conservation of total mechanical energy of the rod, we have
[tex]K_{i} + U_{i} = K_{f} + U_{f}[/tex]
[tex]K_{i} = U_{f} - U_{i}[/tex]
or, mgh = K = 6.86 J
Therefore, h = [tex]\frac{6.86}{mg}[/tex]
= [tex]\frac{0.63}{0.11 \times 9.8}[/tex]
= 0.584 m
Hence, the center of mass rises 0.584 m far above that position.
Answer:
Explanation:
length of rod, L = 0.83 m
mass of rod, M = 110 g = 0.11 kg
angular speed, ω = 5.71 rad/s
Moment of inertia of rod about the fixed point
I = Icm + Mh²
I = 1/12 M L² + M L²/4
I = 1/12 x 0.11 x 0.83 x 0.83 + 0.11 x 0.83 x 0.83 / 4
I = 0.0063 + 0.01895 = 0.0252 kgm²
a) kinetic energy of rod is given by
K = 1/2 Iω²
K = 0.5 x 0.0252 x 5.71 x 5.71 = 0.41 Joule
(b) The kinetic energy at the bottom is equal to the potential energy at a height of the center of mass. Let the centre of mass rises upto height h.
mgh = K
0.11 x 9.8 x h = 0.41
h = 0.381 m
h = 38.1 cm
